What is the solution for 5y^{''}-6y^'+5y=0 ?

The solution for 5y^{''}-6y^'+5y=0 is y=e^{(3t)/5}(c_{1}cos((4t)/5)+c_{2}sin((4t)/5))

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5y^{''}-6y^'+5y=0

5 y^{^{''}} - 6 y^{^{'}} + 5 y = 0

y = e^{\frac{3 t}{5}} (c_{1} cos (\frac{4 t}{5}) + c_{2} sin (\frac{4 t}{5}))

Solution steps

5 y^{''} (t) - 6 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{\frac{3 t}{5}} (c_{1} cos (\frac{4 t}{5}) + c_{2} sin (\frac{4 t}{5}))

y = e^{\frac{3 t}{5}} (c_{1} cos (\frac{4 t}{5}) + c_{2} sin (\frac{4 t}{5}))

f (x) = \frac{16}{x}

\int x^{2} (2 x^{3} - 5) d x

f (x) = sin^{2} (x^{2})

x \to π lim (cot (x))

\frac{d}{d x} (\frac{1}{8 x ^{\frac{7}{8}}})

What is the solution for 5y^{''}-6y^'+5y=0 ?
The solution for 5y^{''}-6y^'+5y=0 is y=e^{(3t)/5}(c_{1}cos((4t)/5)+c_{2}sin((4t)/5))